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Hexagonal Moonshine (1)

Over at the Arcadian Functor, Kea is continuing her series of blog posts on M-theory (the M is supposed to mean either Monad or Motif). A recurrent motif in them is the hexagon and now I notice hexagons popping up everywhere. I will explain some of these observations here in detail, hoping that someone, more in tune with recent technology, may have a use for them.

The three string braid group $B_3 $ is expected to play a crucial role in understanding monstrous moonshine so we should know more about it, for example about its finite dimensional representations. Now, _M geometry_ is pretty good at classifying finite dimensional representations provided the algebra is “smooth” which imples that for every natural number n the variety $\mathbf{rep}_n~A $ of n-dimensional representations of A is a manifold. Unfortunately, the group algebra $A=\mathbb{C} B_3 $ of the three string braid group is singular as we will see in a moment and the hunt for singularities in low dimensional representation varieties will reveal hexagons.

Emil Artin’s three string braid group has two generators $\sigma_1 $ and $\sigma_2 $ (corresponding to one crossing of the first (resp. last) two strings) and one defining equation

$\sigma_1 \sigma_2 \sigma_1 = \sigma_2 \sigma_1 \sigma_2 $

To make the connection with the modular group, it is better to change the generators to $u = \sigma_1 \sigma_2 \sigma_1 $ and $v = \sigma_1 \sigma_2 $ in which case the defining equation becomes $u^2 = v^3 $. Therefore, the variety of all n-dimensional representations of $B_3 $ consists of all pairs of invertible n by n matrices

$~(A,B) \in GL_n(\mathbb{C}) \times GL_n(\mathbb{C})~:~A^2 = B^3 $

which is pretty horrible to work out except for small values of n. For example, when n=1 we get the plane cusp curve $a^2=b^3 $ and while this curve has a unique singular point at (0,0), this point does not correspond to a one-dimensional representation of $B_3 $. Hence, $\mathbf{rep}_1~B_3 $ is the Riemann sphere minus two points ${ 0,\infty } $ and is therefore a manifold. In M-geometry we are not just interested in these representations as points but also in the different ways in which these points can ‘talk to each other’, or more technical, whether these representations can have non-trivial extensions. Recall that an extension of a representation S by a representation S’ is a representation M having S’ as a subrepresentation with quotient-representation S, the trivial extension is the one corresponding to the obvious choice of taking the direct sum of both representations.

Let us work out these extensions for $S=(a,b) $ and $S’=(a’,b’) $ both one-dimensional representations of $B_3 $. An extension of S by S’ must be two-dimensional and is represented by 2×2 matrices of the form

$A = \begin{bmatrix} a & \alpha \\\ 0 & a’ \end{bmatrix}~\quad \text{and} \quad~B=\begin{bmatrix} b & \beta \\\ 0 & b’ \end{bmatrix} $

with $\alpha,\beta \in \mathbb{C} $. Working out the condition that $A^2=B^3 $ leads to the set of equation (using the fact that $a^2=b^3 $ and $~(a’)^2=(b’)^3 $

$ \alpha (a + a’) = \beta (b^2+bb’+(b’)^2) $

This leads to one linear relation between $\alpha $ and $\beta $ unless both coefficients are zero. Further, we have to determine when an extension determined by $~(\alpha,\beta) $ is trivial, that is, isomorphic to $S \oplus S’ $. This translates into matrix-terms to the condition whether $~(A,B) = (L_{\lambda}A_0L_{\lambda}^{-1},L_{\lambda}B_0L_{\lambda}^{-1}) $ for the matrices

$A_0 = \begin{bmatrix} a & 0 \\\ 0 & a’ \end{bmatrix},~\quad B=\begin{bmatrix} b & 0 \\\ 0 & b’ \end{bmatrix}~\quad \text{and} \quad L_{\lambda} = \begin{bmatrix} 1 & \lambda \\\ 0 & 1 \end{bmatrix} $

for some $\lambda \in \mathbb{C} $. This leads to the additional equations $\alpha = \lambda(a’-a) $ and $\beta = \lambda(b’-b) $.

Summarizing we have that there are NO extensions of S by S’ unless either S=S’ (in which case the isomorphism condition gives no extra condition and hence there is a one-dimensional family of self-extensions of S by itself for every S) or $S \not= S’ $ and the two coefficients of the linear relation between $\alpha $ and $\beta $ are both zero (in which case there is again a one-dimensional family of extensions of S by S’). The condition on S and S’ for both coefficients to vanish is that

$a’=-a $ and either $b’=\zeta b $ or $b’ = \zeta^2 b $

for $\zeta $ a third root of unity. Therefore, each one-dimensional $B_3 $-representation $S_1=(a,b) $ (a point on the Riemann sphere minus $0,\infty $) belongs to a unique clan of six one-dimensional representations, having as the other members the representations

$S_2 = (-a,\zeta b),~S_3=(a,\zeta^2 b),~S_4=(-a,b),~S_5=(a,\zeta b),~S_6=(-a,\zeta^2 b) $

Different clans don’t talk to each other and within a clan the lines of communication (the spaces of extensions) can be depicted by the following hexagonal quiver

[tex]\xymatrix{& \vtx{S_1} \ar@(ul,ur) \ar@/^/[dl] \ar@/^/[dr] & \\\ \vtx{S_6} \ar@(dl,ul) \ar@/^/[ur] \ar@/^/[d] & & \vtx{S_2} \ar@(ur,dr) \ar@/^/[ul] \ar@/^/[d] \\\ \vtx{S_5} \ar@(dl,ul) \ar@/^/[u] \ar@/^/[dr] & & \vtx{S_3} \ar@(ur,dr) \ar@/^/[u] \ar@/^/[dl] \\\ & \vtx{S_4} \ar@(dr,dl) \ar@/^/[ur] \ar@/^/[ul] & }[/tex]

Having fully analyzed the one-dimensional situation, let’s turn to $\mathbf{rep}_2~B_3 $ and see that it has singularities and hence that $B_3 $ is not smooth in M-geometry. The three string braid group has a 2-parameter family of simple 2-dimensional representation, whence the component of the representation variety $\mathbf{rep}_2~B_3 $ containing the simples is 5-dimensional. Hence, if this component would be a manifold, the tangent space in every point should be of dimension 5. If M is a semi-simple n-dimensional representation, then the tangent space in M to $\mathbf{rep}_n~B_3 $ is known to be of dimension $n^2 – dim~Aut(M) + dim~Ext(M,M) $ where $Ext(M,M) $ is the space of self-extensions and $Aut(M) $ is the automorphism group of M.

Apply this to a semi-simple 2-dimenional representation $M=S \oplus S’ $ where S and S’ determine an edge of an hexagon in a clan. Then,

$Ext(M,M) = Ext(S,S) \oplus Ext(S,S’) \oplus Ext(S’,S) \oplus Ext(S’,S’) $

which by the above calculations is a 4-dimensional space. Further, S being non-isomorphic to S’ it follows from Schur’s lemma that the automorphism group of M is two-dimensional, totting up to a 6-dimensional tangent space… so M must be a singularity!

In fact, one can show that the points in the irreducible component of two-dimensional $B_3 $ representations containing the simples are singular if and only if the representation is semi-simple $M = S \oplus S’ $ with the components either a vertex of a hexagon (that is S=S’) or an edge of an hexagon. In this way, the strange hexagonal clans of one-dimensional braid-representations are the source of singularities for $B_3 $.

In _another post_ we will see that the hexagonal pattern also pops up in studying representations of the modular group $PSL_2(\mathbb{Z}) $ and the extended modular group $PGL_2(\mathbb{Z}) $. The latter case is pretty interesting as we will encounter a Moebius band made of two hexagons…

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